Question

2HI -> H2 I2

Keq= 49.5 at 440'C

If 5 0. M HI is initially placed into a container, what will be the equilibrium [ ] HI ?

Answer 1

$K_{eq}=\frac{[H_2][I_2]}{[HI]^2}$

$2HI \to H_2+I_2\\$

$5-x$ $\frac{x}{2}$ $\frac{x}{2}$

The initian concentration is $5 M.$

$K_{eq}=49.5$ $=\frac{\frac{x}{2}\times\frac{x}{2}}{(5-x)^2}$

$\implies197x^2-1980x+4950=0$

$\implies x=4.68$

The equilibrium concentration of $[HI]$ is $=5-x=5-4.68=0.32\ M$